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Ionic Equilibrium question

2007 · Shift 0 · Q6
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  5. /2007 · Shift 0 · Q6

Ionic Equilibrium question

2007 · Shift 0 · Q6

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The first and second dissociation constants of an acid H2AH_2AH2​A are 1.0 ×\times× 10−5 and 5.0 ×\times× 10−10 respectively. The overall dissociation constant of the acid will be :
  1. A
    5.0 ×\times× 10−5
  2. B
    5.0 ×\times× 1015
  3. C
    5.0 ×\times× 10−15
  4. D
    5.0 ×\times× 105
View written solutionFree

Correct answer: C

  1. Write the two stepwise dissociations of the diprotic acid

For the acid H2AH_2AH2​A:

H2A⇌H++HA−H_2A \rightleftharpoons H^+ + HA^-H2​A⇌H++HA−

Its first dissociation constant is:

K1=1.0×10−5K_1 = 1.0 \times 10^{-5}K1​=1.0×10−5

The second dissociation is:

HA−⇌H++A2−HA^- \rightleftharpoons H^+ + A^{2-}HA−⇌H++A2−

Its second dissociation constant is:

K2=5.0×10−10K_2 = 5.0 \times 10^{-10}K2​=5.0×10−10

  1. Overall dissociation reaction

Adding the two steps:

H2A⇌2H++A2−H_2A \rightleftharpoons 2H^+ + A^{2-}H2​A⇌2H++A2−

The overall dissociation constant is the product of the stepwise constants:

K=K1K2K = K_1 K_2K=K1​K2​

  1. Calculate the value

K=(1.0×10−5)(5.0×10−10)K = (1.0 \times 10^{-5})(5.0 \times 10^{-10})K=(1.0×10−5)(5.0×10−10)

Multiply coefficients:

1.0×5.0=5.01.0 \times 5.0 = 5.01.0×5.0=5.0

Add exponents:

10−5×10−10=10−1510^{-5} \times 10^{-10} = 10^{-15}10−5×10−10=10−15

So,

K=5.0×10−15K = 5.0 \times 10^{-15}K=5.0×10−15

  1. Match with the options

The correct option is:

C: 5.0×10−15\boxed{\text{C: } 5.0 \times 10^{-15}}C: 5.0×10−15​

  1. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So, they agree.

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